The problem is :
Create an array of N integers. Ask from the user how many element he/she wants to input in the array. Determine the sum of all numbers, the average of all even numbers and count how many are odd numbers and how many zero's were inputed ? at the end, print the accumulated and counted values of the array.
Karon akong problem man gud kay kung mu input ko ug
6 elements din mo-input ko ug numbers nga
0,1,2,3,4,5 kay ang iyahang average sa even numbers kay
2.5 instead of
2 kay ang zero kay even mana diba? so dapat jud ana kay (0+2+4)/3 kay 3 man ka even numbers ang na input. Confused au ko ngano 2.5 jud iyahang output sa average sa even.
Here is my code :
Code:
#include<stdio.h>
#include<conio.h>
main()
{
int num[]={};
int max=0,i,sumofzero=0,zero=0,sumodd=0,sumeven=0;
float sum=0;
float ave=0;
printf("How many elements you want? : ");
scanf("%d", &max);
for(i=0;i<max;i++)
{
printf("Enter number %d : \n",i );
scanf("%d", &num[i]);
sum+=num[i];
}
for(i=0;i<max;i++)//for even average numbers
{
if(num[i]%2==0)
ave=(sum/max);
}
for(i=0;i<max;i++)//for odd numbers
{
if(num[i]%2!=0)
sumodd+=1;
}
for(i=0;i<max;i++)//for zero numbers
{
if(num[i]==0)
sumofzero+=num[i]+1;
}
printf("The sum of all numbers is : %.2f \n", sum);
printf("The average of all even numbers is %.2f \n", ave);
printf("The total odd numbers were inputed is : %d \n", sumodd);
printf("The sum of all zero numbers is : %d \n", sumofzero);
getch();
}
pero if mu input ko ug 5 ra ka elements din ang numbers kay
0,1,2,3,4 kay ang output sa average is
2 w/c is sakto. Unsa diay kulang or sayop sa akong code? Hope maka tabang mo, thank you!